Problem: What is the slope of the line tangent to $f(x) = -x^{2}+2x+6$ at $x = -1$ ?
Answer: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-(x+h)^{2}+2(x+h)+6) - (-x^{2}+2x+6)}{h}$ $ = \lim_{h \to 0} \frac{(-(x^{2}+2x h+h^{2})+2(x+h)+6) - (-x^{2}+2x+6)}{h}$ $ = \lim_{h \to 0} \frac{-x^{2}-2(x h)-h^{2}+2x+2h+6+x^{2}-2x-6}{h}$ $ = \lim_{h \to 0} \frac{-2(x h)-h^{2}+2h}{h}$ $ = \lim_{h \to 0} -2x-h+2$ $ = -2x+2$ $ = (-2)(-1)+2$ $ = 4$